17
Ionic Equilibria III:
The Solubility
Product Principle
1
Determination of Solubility
Product Constants
Example 17-7: One liter of saturated silver chloride
solution contains 0.00192 g of dissolved AgCl at
25oC. Calculate the molar solubility of, and Ksp for,
AgCl.
• The molar solubility can be easily calculated
from the data:
? mol AgCl 0.00192 g AgCl 1 mol AgCl


L
L
143 g AgCl
 5 mol AgCl
 1.34  10
L
2
Determination of Solubility
Product Constants
Example 17-8: One liter of saturated calcium
fluoride solution contains 0.0167 gram of CaF2 at
25oC. Calculate the molar solubility of, and Ksp for,
CaF2.
1. Calculate the molar solubility of CaF2.
? mol CaF2 0.0167 g CaF2 1 mol


L
1.0 L
78.1 g
 4 mol CaF2
 2.14  10
L
3
Determination of Solubility
Product Constants
• From the molar solubility, we can find the ion
concentrations in saturated CaF2. Then use
those values to calculate the Ksp.
– Note: You are most likely to leave out the factor of 2
for the concentration of the fluoride ion!
CaF2
CaF
2


2 +2 
- F1
Ca

2

Ca aq  + 2 F aq 
4
4 )
2.14  10  4 4M 2.14  10  4M
2(2.14  10  4M
2.14  10 M 2.14  10 M 2(2.14  10 M )

F 
 2.14  10 4.28 10 
K sp  Ca
2
 2
4
4 2
 3.92  10 11
4
Uses of Solubility
Product Constants
Example 17-9: Calculate the molar solubility of
barium sulfate, BaSO4, in pure water and the
concentration of barium and sulfate ions in
saturated barium sulfate at 25oC. For barium
sulfate, Ksp= 1.1 x 10-10.
5
Uses of Solubility
Product Constants
2+
2

BaSO4 s  Ba  aq   SO4 aq 
xM
 xM
xM

Ksp  Ba
2

2
SO4
  11.  10
10
6
Uses of Solubility
Product Constants
• Make the algebraic substitution of x’s into
solubility product expression and solve for x,
giving the ion concentrations.
x x   1.110
10
5
x  1.0 10 M
Ba   SO   1.0 10
2
2
4
5
M
7
Uses of Solubility
Product Constants
• Finally, to calculate the mass of BaSO4 in 1.00 L
of saturated solution, use the definition of
molarity.
5
? g BaSO 4 1.0 10 mol 234 g


L
L
mol
3 g BaSO 4
 2.3 10
L
8
Uses of Solubility
Product Constants
Example 17-10: The solubility product constant for
magnesium hydroxide, Mg(OH)2, is 1.5 x 10-11.
Calculate the molar solubility of magnesium
hydroxide and the pH of a saturated magnesium
hydroxide solution at 25oC.
9
Uses of Solubility
Product Constants
• Be careful, do not forget the stoichiometric
coefficient of 2!
2


Mg(OH) 2  Mg aq   2 OH  aq 
xM

 xM
Ksp  Mg
2
2 xM
OH 
 2
10
Uses of Solubility
Product Constants
• Substitute the algebraic expressions into the
solubility product expression.
2
11
.  10
 x 2 x   15
4 x  15
.  10
3
11
x  3.75  10
3
x  16
.  10
4
12
 molar solubility
11
Uses of Solubility
Product Constants
• Solve for the pOH and pH.
.  1011
 x 2 x   15
2
4 x 3  15
.  1011
x  3.75  10
3
x  16
.  10
4
12
 molar solubility
OH   2  M  3.2  10
-
4
M
pOH  3.49 pH  10.51
12
The Common Ion Effect in
Solubility Calculations
Example 17-11: Calculate the molar solubility of
barium sulfate, BaSO4, in 0.010 M sodium sulfate,
Na2SO4, solution at 25oC. Compare this to the
solubility of BaSO4 in pure water. (Example 20-3).
(What is the common ion? How was a common
ion problem solved in Chapter 19?)
13
The Common Ion Effect in
Solubility Calculations
1. Write equations to represent the equilibria.
Na 2SO 4 
 2 Na  SO
100%
0.010M
BaSO 4s 
xM


+
2
4
2(0.010M ) (0.010M )
Ba
xM
2
aq 
 SO
2
4 aq 
xM
14
The Common Ion Effect in
Solubility Calculations
2. Substitute the algebraic representations of the
concentrations into the Ksp expression and
solve for x.



K sp  Ba 2 SO 24  1.1 10 10
  x 0.010  x 
The simplifying assumption can be applied.
0.010  x   0.010
0.010 x = 1.1 10 -10
x  1.1 10 8  molar solubility of BaSO 4
15
The Reaction Quotient in
Precipitation Reactions
Example 17-12: We mix 100 mL of 0.010 M
potassium sulfate, K2SO4, and 100 mL of 0.10 M
lead (II) nitrate, Pb(NO3)2 solutions. Will a
precipitate form?
16
The Reaction Quotient in
Precipitation Reactions
1. Write out the solubility expressions.
K2SO4
H 2O 100%
 
 2 K 
+
2
SO4
Pb NO3 2  
 Pb  2
H 2O 100%
2+
NO3
Will PbSO 4 precipitate?
17
The Reaction Quotient in
Precipitation Reactions
• Calculate the Qsp for PbSO4.
– Assume that the solution volumes are additive.
– Concentrations of the important ions are:
MPbPb2+
M
2+
MSO24
100
100mL
mL 0.10
0.10MM
2+2+
.050MMPbPb

0.0050
200mL
mL
200
100 mL  0.010 M
2
 0.0050 M SO 4
200 mL
18
The Reaction Quotient in
Precipitation Reactions
• Finally, calculate Qsp for PbSO4 and compare it
to the Ksp.

Qsp  Pb
2

2
SO4

 0.050 0.0050
 2.5  10
Ksp  18
.  10
4
8
for PbSO4
Qsp  Ksp therefore solid forms
19
The Reaction Quotient in
Precipitation Reactions
Example 17-13: What concentration of sulfide ions,
from a soluble compound such as Na2S, is required
to reduce the Hg2+ concentration to 1.0 x 10-8 M?
For HgS, Ksp=3.0 x 10-53.
 HgS
Hg 2  S2 

 
solve for S 
K sp  Hg 2 S2  3.0 10 53
2
S   Hg 
2
K sp
2
3.0 10 53
 45


3
.
0

10
M
8
1.0 10
If enough S2- , in the form of Na 2S, is added to just
slightly exceed 3.0 10
 45
M the mercury wi ll precipitate.
20
The Reaction Quotient in
Precipitation Reactions
Example 17-14: Refer to example 16-13. What
volume of the solution (1.0 x 10-8 M Hg2+ ) contains
1.0 g of mercury?
2
1 mol Hg
1.0 L
? L  1.0 g Hg 

2
8
2
201 g Hg 1.0  10 mol Hg
2
 5.0  105 L  125,000 gal
21
Fractional Precipitation
Example 17-15: If solid sodium chloride is slowly
added to a solution that is 0.010 M each in Cu+, Ag+,
and Au+ ions, which compound precipitates first?
Calculate the concentration of Cl- required to initiate
precipitation of each of these metal chlorides.
AuCl has the smallest K sp so it precipitates first.
By the same reasoning, CuCl will precipitate last.
1. Calculate the concentrat ion of Cl - required to precipitate AuCl.
22
Fractional Precipitation
AuCl has the smallest K sp so it precipitates first.
By the same reasoning, CuCl will precipitate last.
1. Calculate the concentrat ion of Cl - required to precipitate AuCl.
Au Cl   2.0 10

 

13
 K sp
13
13
2
.
0

10
2
.
0

10
11
Cl  


2
.
0

10
M

Au
0.010
 
23
Fractional Precipitation
• Repeat the calculation for silver chloride.
 Cl   1.8 10
K sp  Ag
 

1.8  10
Cl 

Ag


10
10
10
 
1.8 10

0.010
8
 1.8  10 M
24
Fractional Precipitation
• Finally, for copper (I) chloride to precipitate.
  

7
7
7
K sp  Cu Cl  1.9 10
+
 
1.9 10
Cl 
+
Cu

 
1.9 10

0.010
5
 1.9 10 M
25
Fractional Precipitation
Example 17-16: Calculate the percentage of Au+
ions that precipitate before AgCl begins to
precipitate.
– Use the [Cl-] from Example 16-15 to determine the
[Au+] remaining in solution just before AgCl begins to
precipitate. Au  Cl   2.0  1013
  
2.0  10
Au   Cl 
13




2.0  1013

18
.  108


Au   11
.  105  Au  unprecipitated
26
Fractional Precipitation
• The percent of Au+ ions unprecipitated just
before AgCl precipitates is:

Au 

Au 

% Au
+
unprecipitated
unprecipitated

original
 100%
5
1.1 10

 100  0.1% unprecipitated
0.010
• Therefore, 99.9% of the Au+ ions precipitates
before AgCl begins to precipitate.
27
Fractional Precipitation
• A similar calculation for the concentration
of Ag+ ions unprecipitated before CuCl
begins to precipitate is:
Ag Cl   1.8 10
1.8 10
1.8 10
Ag   Cl   1.9 10
Ag   9.5 10  Ag  unprecipitated


10
10

10


5
6

28
Fractional Precipitation
• The percent of Ag+ ions unprecipitated just
before AgCl precipitates is:

Ag 

Ag 

% Ag
+
unprecipitated
unprecipitated

original
100%
6
9.5 10

100  0.095% unprecipitated
0.010
• Thus, 99.905% of the Ag+ ions precipitates
before CuCl begins to precipitate.
29
Simultaneous Equilibria Involving
Slightly Soluble Compounds
Example 17-17: If 0.10 mole of ammonia and 0.010
mole of magnesium nitrate, Mg(NO3)2, are added to
enough water to make one liter of solution, will
magnesium hydroxide precipitate from the solution?
–
For Mg(OH)2, Ksp = 1.5 x 10-11. Kb for NH3 = 1.8 x 10-5.
1.Calculate Qsp for Mg(OH)2 and compare it to Ksp.
–
–
Mg(NO3)2 is a soluble ionic compound so [Mg2+] = 0.010 M.
Aqueous ammonia is a weak base that we can calculate
[OH-].
30
Simultaneous Equilibria Involving
Slightly Soluble Compounds

- 

 
NH

H
O

NH
OH

+
H
O

NH
OH
3
2
4
NH 3 H O
2  NH 4OH
3
2
4
10
xMxMxM
xMxM
00..10
10xxxMMM xM



NH
OH




NH
OH
5
4
K


1
.
8

10
K b  NH 
 1.8 10
NH 3 

x  x 




x
x
0.10  x 


0
.
10

x
x  1.8 10  x = 1.3 10 M  OH 

4

5
b
3
2
6
-3

31
Simultaneous Equilibria Involving
Slightly Soluble Compounds
• Once the concentrations of both the
magnesium and hydroxide ions are
determined, the Qsp can be calculated and
compared to the Ksp.
2
 2

OH 
 0.010 1.3  10 
Q sp  Mg
3 2
 1.7  10
8
Q sp  K sp , thus Mg(OH) 2 will precipitate.
32
Simultaneous Equilibria Involving
Slightly Soluble Compounds
Example 17-18: How many moles of solid
ammonium chloride, NH4Cl, must be used to prevent
precipitation of Mg(OH)2 in one liter of solution that is
0.10 M in aqueous ammonia and 0.010 M in
magnesium nitrate, Mg(NO3)2 ? (Note the similarity
between this problem and Example 16-17.)
•Calculate the maximum [OH-] that can exist in a
solution that is 0.010 M in Mg2+.
33
Simultaneous Equilibria Involving
Slightly Soluble Compounds

Ksp  Mg
OH 
 2
2
OH 
 2
11
15
.  10

2
Mg

 15
.  10
11
15
.  10

0.010
9
OH   3.9  10


 15
.  10
11
5
M
34
Simultaneous Equilibria Involving
Slightly Soluble Compounds
• Using the maximum [OH-] that can exist in
solution, determine the number of moles of NH4Cl
required to buffer 0.10 M aqueous ammonia so
that the [OH-] does not exceed 3.9 x 10-5 M.
NH 4Cl 100%
 NH 4  Cl 
xM
NH 3 +
xM
+

H 2O  NH 4 + OH
.  3.9  10  M
010
-5
3.9  10 M 3.9  10 M
-5
-5
35
Simultaneous Equilibria Involving
Slightly Soluble Compounds
NH 4 Cl 100%

 NH 4  Cl 
xM
 NH 
H 2O 
4
NH 3 
0.10  3.9 10 M 3.9 10

NH OH 
K =
 1.8 10
-5
+
4
b
xM


-5
OH -
M 3.9 10-5 M
5
NH 3 

x  3.9 10 3.9 10 

0.10  3.9 10 
-5
-5
-5
The simplifying assumption can be applied.
36
Simultaneous Equilibria Involving
Slightly Soluble Compounds
3.9 10 x  1.8 10
-5
5
0.10
x  0.046 M  NH 4 Cl 
Because there is 1.0 L of solution,
there are 0.046 mol NH 4 Cl.
37
Simultaneous Equilibria Involving
Slightly Soluble Compounds
• Check these values by calculating Qsp for
Mg(OH)2.

OH 
 0.0103.9 10 
Q sp = Mg
2+
 2
5 2
 1.5 10 11
Q sp  K sp
Thus this system is at equilibrium!
38
Simultaneous Equilibria Involving
Slightly Soluble Compounds
• Use the ion product for water to calculate
the [H+] and the pH of the solution.
 

10
.  10
 H   OH 
H + OH   10
.  10-14
-14
+

10
.  10
10
H 

2
.
6

10
M
5
3.9  10
pH  9.59
 
-14
+
39
Complex Ion Equilibria
Example 17-19: How many moles of ammonia must
be added to 2.00 L of water so that it will just
dissolve 0.010 mole of silver chloride, AgCl?
• The reaction of interest is:

AgCl s 

2 NH 3 
Ag(NH 3 ) 2 + Cl 5.0 10 3 M 2 (5.0 10 3 ) M
5.0 10 3 M
5.0 10 3 M
The actual reaction ratios have been used for these values.
40
Complex Ion Equilibria
• Two equilibria are involved when silver chloride
dissolves in aqueous ammonia.
AgCl 
 Ag   Cl 
Ag(NH 3 ) 2
 Ag   2 NH

3
  
Ag  NH 


 6.3  10
Ag(NH ) 
Ksp  Ag  Cl   18
.  1010

Kd
2
3

3 2
10
41
Complex Ion Equilibria
• The [Ag+] in the solution must satisfy both
equilibrium constant expressions. Because the
[Cl-] is known, the equilibrium concentration of
Ag+ can be calculated from Ksp for AgCl.
Ag Cl   1.8 10
1.8  10
Ag   Cl 


10
10


42
Complex Ion Equilibria
Ag Cl   1.8 10
1.8 10
Ag   Cl 


10
10


10
1.8 10

5.0 10 3
8
+
 3.6 10 M Ag
 
+
Which is the maximum Ag possible.
43
Complex Ion Equilibria
• Substitute the maximum [Ag+] into the dissociation
constant expression for [Ag(NH3)2]+ and solve for
the equilibrium concentration of NH3.

Ag2NH 

3
K    8.75
 6.3
 10
NH
10
3Ag(NH ) 

3.6 10 NH 

NH
  0.094M

d
2
8
3

3 2
8
2
3
3 5.0 10 3
44
Complex Ion Equilibria
• The amount just calculated is the equilibrium
concentration of NH3 in the solution. But the
total concentration of NH3 is the equilibrium
amount plus the amount used in the complex
formation.
NH 3 total  0.094M  2(5.0 10
3
) M  0.104 M
equilibrium
reaction
amount
amount
45
Complex Ion Equilibria
• Finally, calculate the total number of moles of
ammonia necessary.
0.104 mol NH 3
? mol NH 3  2.0L 
L
 0.21 M
46
Synthesis Question
Most kidney stones are made of calcium oxalate,
Ca(O2CCO2). Patients who have their first kidney
stones are given an extremely simple solution to
stop further stone formation. They are told to drink
six to eight glasses of water a day. How does this
stop kidney stone formation?
47
Synthesis Question
The solubility product expression
for calcium oxalate is
2
2

Ca(COO) 2 s   H 2 O  Ca aq   COO 2aq 
Drinking more water shifts the equilibrium
to the right, towards ions in solution
rather tha n as a solid stone.
48